Let $S$ be a surface in 3D described by the equation $\sqrt{x^2 + y^2} - z^2 = 0$. What is the equation of the plane tangent to $S$ at $(3, 4, \sqrt{5})$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{3}{5} (x - 3) - \dfrac{4}{5}(y - 4) - \sqrt{5}(z - \sqrt{5}) = 0$ (Choice B) B $\dfrac{4}{5} (x + 3) + \dfrac{3}{5}(y + 4) - 2\sqrt{5}(z + \sqrt{5}) = 0$ (Choice C) C $\dfrac{3}{5} (x - 3) + \dfrac{4}{5}(y - 4) - 2\sqrt{5}(z - \sqrt{5}) = 0$ (Choice D) D $\dfrac{4}{5} (x - 3) + \dfrac{3}{5}(y - 4) - 2\sqrt{5}(z - \sqrt{5}) = 0$
Explanation: The equation for a tangent plane of an implicitly defined surface $F(x, y, z) = 0$ at the point $(a, b, c)$ is: $F_x(x - a) + F_y(y - b) + F_z(z - c) = 0$ [What's the intuition behind the formula?] Let's find $F_x$, $F_y$, and $F_z$. $\begin{aligned} F_x &= \dfrac{2x}{2\sqrt{x^2 + y^2}} = \dfrac{3}{5}\\ \\ F_y &= \dfrac{2y}{2\sqrt{x^2 + y^2}} = \dfrac{4}{5} \\ \\ F_z &= -2z = -2\sqrt{5} \end{aligned}$ Putting it all together, here's the equation for the tangent plane of $S$ at $(3, 4, \sqrt{5})$ : $\dfrac{3}{5} (x - 3) + \dfrac{4}{5}(y - 4) - 2\sqrt{5}(z - \sqrt{5}) = 0$